Python program to check if any key has all the given list elements
Last Updated :
26 Feb, 2023
Given a dictionary with list values and a list, the task is to write a Python program to check if any key has all the list elements.
Examples:
Input : test_dict = {‘Gfg’ : [5, 3, 1, 6, 4], ‘is’ : [8, 2, 1, 6, 4], ‘best’ : [1, 2, 7, 3, 9], ‘for’ : [5, 2, 7, 8, 4, 1], ‘all’ : [8, 5, 3, 1, 2]}, find_list = [7, 9, 2]
Output : True
Explanation : best has all values, 7, 9, 2 hence 2 is returned.
Input : test_dict = {‘Gfg’ : [5, 3, 1, 6, 4], ‘is’ : [8, 2, 1, 6, 4], ‘best’ : [1, 2, 7, 3, 19], ‘for’ : [5, 2, 7, 8, 4, 1], ‘all’ : [8, 5, 3, 1, 2]}, find_list = [7, 9, 2]
Output : False
Explanation : No list has all values as find list.
Method #1 : Using issuperset() + loop
In this, we perform the task of finding the presence of all elements in the values list using issuperset(), and a loop is used to iterate all the keys.
Python3
test_dict = {'Gfg': [5, 3, 1, 6, 4],
'is': [8, 2, 1, 6, 4],
'best': [1, 2, 7, 3, 9],
'for': [5, 2, 7, 8, 4, 1],
'all': [8, 5, 3, 1, 2]}
print("The original dictionary is : " + str(test_dict))
find_list = [7, 9, 2]
res = False
for key in test_dict:
if set(test_dict[key]).issuperset(find_list):
res = True
print("Is any value list superset ? : " + str(res))
|
Output:
The original dictionary is : {‘Gfg’: [5, 3, 1, 6, 4], ‘is’: [8, 2, 1, 6, 4], ‘best’: [1, 2, 7, 3, 9], ‘for’: [5, 2, 7, 8, 4, 1], ‘all’: [8, 5, 3, 1, 2]}
Is any value list superset ? : True
Time complexity: O(mn), where m is the number of keys in the dictionary and n is the length of the longest value list in the dictionary.
Auxiliary space: O(n)
The auxiliary space complexity of the given code is O(n), where n is the length of the find_list.
Method #2 : Using any() + issuperset()
In this, we perform the task of checking for all keys using any(). Rest all functions are similar to the above method.
Python3
test_dict = {'Gfg': [5, 3, 1, 6, 4],
'is': [8, 2, 1, 6, 4],
'best': [1, 2, 7, 3, 9],
'for': [5, 2, 7, 8, 4, 1],
'all': [8, 5, 3, 1, 2]}
print("The original dictionary is : " + str(test_dict))
find_list = [7, 9, 2]
res = any(set(sub).issuperset(find_list)
for sub in test_dict.values())
print("Is any value list superset ? : " + str(res))
|
Output:
The original dictionary is : {‘Gfg’: [5, 3, 1, 6, 4], ‘is’: [8, 2, 1, 6, 4], ‘best’: [1, 2, 7, 3, 9], ‘for’: [5, 2, 7, 8, 4, 1], ‘all’: [8, 5, 3, 1, 2]}
Is any value list superset ? : True
Method #3 : Without issuperset()
Python3
def contains(x,y):
c=0
for i in y:
if i in x:
c+=1
if(c==len(y)):
return True
return False
test_dict = {'Gfg': [5, 3, 1, 6, 4],
'is': [8, 2, 1, 6, 4],
'best': [1, 2, 7, 3, 9],
'for': [5, 2, 7, 8, 4, 1],
'all': [8, 5, 3, 1, 2]}
print("The original dictionary is : " + str(test_dict))
find_list = [7, 9, 2]
res = False
for i in test_dict.keys():
if(contains(test_dict[i],find_list)):
res=True
break
print("Is any value list superset ? : " + str(res))
|
Output
The original dictionary is : {'Gfg': [5, 3, 1, 6, 4], 'is': [8, 2, 1, 6, 4], 'best': [1, 2, 7, 3, 9], 'for': [5, 2, 7, 8, 4, 1], 'all': [8, 5, 3, 1, 2]}
Is any value list superset ? : True
Using Recursion:
Python3
def contains(x,y,res):
if len(y) == 0:
return res
if y[0] in x:
return contains(x,y[1:],True)
else:
return contains(x,y[1:],False)
def recursion(test_dict, find_list, res):
if len(test_dict) == 0:
return res
key, value = test_dict.popitem()
if contains(value, find_list,False):
res = True
return recursion(test_dict, find_list, res)
test_dict = {'Gfg': [5, 3, 1, 6, 4],
'is': [8, 2, 1, 6, 4],
'best': [1, 2, 7, 3, 9],
'for': [5, 2, 7, 8, 4, 1],
'all': [8, 5, 3, 1, 2]}
print("The original dictionary is : " + str(test_dict))
find_list = [7, 9, 2]
res = recursion(test_dict, find_list, False)
print("Is any value list superset ? : " + str(res))
|
Output
The original dictionary is : {'Gfg': [5, 3, 1, 6, 4], 'is': [8, 2, 1, 6, 4], 'best': [1, 2, 7, 3, 9], 'for': [5, 2, 7, 8, 4, 1], 'all': [8, 5, 3, 1, 2]}
Is any value list superset ? : True
Time Complexity: O(N*M) where N is the number of keys in the dictionary, and M is the length of the find_list.
Auxiliary Space: O(n)
Method #5 : Using filter() and lambda functions:
This code demonstrates how to check if any of the values (lists) in a given dictionary are supersets of a target list.
The given dictionary “test_dict” contains keys (strings) and corresponding values (lists of integers). The target list “find_list” contains some integers that we want to find in any of the value lists of the dictionary.
The code first prints the original dictionary. It then applies the filter function to the dictionary values. The filter function takes a lambda function as an argument which checks if a given set of integers (obtained from the dictionary values) is a superset of the target list. The filter function returns a filter object which is then passed to the any function. The any function returns True if any of the elements in the filter object are True, meaning that at least one value list in the dictionary is a superset of the target list.
The code then prints the result indicating whether any value list in the dictionary is a superset of the target list or not.
Python3
test_dict = {'Gfg': [5, 3, 1, 6, 4],
'is': [8, 2, 1, 6, 4],
'best': [1, 2, 7, 3, 9],
'for': [5, 2, 7, 8, 4, 1],
'all': [8, 5, 3, 1, 2]}
find_list = [7, 9, 2]
print("The original dictionary is : " + str(test_dict))
res = any(filter(lambda x: set(x).issuperset(find_list), test_dict.values()))
print("Is any value list superset ? : " + str(res))
|
Output
The original dictionary is : {'Gfg': [5, 3, 1, 6, 4], 'is': [8, 2, 1, 6, 4], 'best': [1, 2, 7, 3, 9], 'for': [5, 2, 7, 8, 4, 1], 'all': [8, 5, 3, 1, 2]}
Is any value list superset ? : True
Time Complexity: O(n*m)
Auxiliary Space: O(1)
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